An Alaskan rescue plane traveling 37 m/s
drops a package of emergency rations from
a height of 149 m to a stranded party of
explorers.
The acceleration of gravity is 9.8 m/s
2
.
Where does the package strike the ground
relative to the point directly below where it
was released?
Answer in units of m
007 (part 2 of 3) 10.0 points
What is the horizontal component of the velocity just before it hits?
Answer in units of m/s
What is the vertical component of the velocity
just before it hits? (Choose upward as the
positive vertical direction)
Answer in units of m/s
To find where the package strikes the ground relative to the point directly below where it was released, we need to determine the time it takes for the package to fall. We can use the equation:
Δy = V₀yt + 1/2gt^2
Where:
Δy = 149 m (height of the drop)
V₀y = 0 m/s (initial vertical velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
t = time
We can rearrange the equation to solve for t:
t = sqrt(2Δy / g)
Substituting the given values, we get:
t = sqrt(2 * 149 m / 9.8 m/s^2)
t ≈ 5.01 s
Now, to find the horizontal component of the velocity just before it hits, we can use the equation:
Vx = V₀x
Where:
Vx = horizontal component of velocity
V₀x = 37 m/s (given velocity of the rescue plane)
The horizontal component of velocity remains constant throughout the motion, so it is equal to the initial horizontal velocity of the package. Therefore:
Vx = V₀x = 37 m/s
To find the vertical component of the velocity just before it hits, we can use the equation:
Vy = V₀y + gt
Where:
Vy = vertical component of velocity
V₀y = 0 m/s (initial vertical velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
t = 5.01 s (time calculated earlier)
Substituting the given values, we get:
Vy = 0 m/s + (9.8 m/s^2)(5.01 s)
Vy ≈ 49 m/s
Therefore, the answer is:
- The package strikes the ground 149 m below the point directly below where it was released.
- The horizontal component of the velocity just before it hits is 37 m/s.
- The vertical component of the velocity just before it hits is approximately 49 m/s.
To find the answer, we can use the equations of motion for an object in free fall.
1. The time taken for the package to reach the ground can be found using the equation:
h = (1/2) * g * t^2
rearranging the equation, we get:
t = sqrt((2h) / g)
where h is the height (149 m) and g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values, we get:
t = sqrt((2 * 149) / 9.8) ≈ 5.07 s
2. The horizontal distance the package travels can be found using the equation:
d = v * t
where v is the horizontal velocity (same as the plane's velocity) and t is the time of flight.
Plugging in the values, we get:
d = 37 m/s * 5.07 s ≈ 187.59 m
Therefore, the package strikes the ground approximately 187.59 meters away from the point directly below where it was released.
3. The horizontal component of the velocity just before it hits is the same as the plane's velocity, which is 37 m/s.
4. The vertical component of the velocity just before it hits can be found using the equation:
v = g * t
where g is the acceleration due to gravity and t is the time of flight.
Plugging in the values, we get:
v = 9.8 m/s^2 * 5.07 s ≈ 49.686 m/s
Therefore, the vertical component of the velocity just before it hits is approximately 49.686 m/s, in the upward direction.